1. ## Implicit Differentiation #3

Find dy/dx for the following:

y^3=x^3/(2-xy)

This is my work.. can someone check to see if I have the correct answer and method? Thanks

d/dx (3y^2)=(3x^2)(2-xy)-x^3[(-1)y+x(1)dy/dx]/(2-xy)^2

d/dx (3y^2)=(3x^2)(2-xy)+x^3y-x(dy/dx)]/(2-xy)^2

(2-xy)^2(3y^2)(dy/dx)+x (dy/dx)=3x^2(2-xy)+x^3y
dy/dx[(2-xy)^2(3y^2)+x]=3x^2(2-xy)+x^3y
dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)+x]

2. I see a mistake on the RHS on the first step:

You have $\displaystyle \frac{(2-xy)(3x^2) - x^3(-y+xdy/dx)}{(2-xy)^2}$

Check the signs again, I believe it should be: $\displaystyle \frac{(2-xy)(3x^2) - x^3(-y-xdy/dx)}{(2-xy)^2}$

3. Originally Posted by eXist
I see a mistake on the RHS on the first step:

You have $\displaystyle \frac{(2-xy)(3x^2) - x^3(-y+xdy/dx)}{(2-xy)^2}$

Check the signs again, I believe it should be: $\displaystyle \frac{(2-xy)(3x^2) - x^3(-y-xdy/dx)}{(2-xy)^2}$

d/dx (3y^2)=(3x^2)(2-xy)-x^3[(-1)y-x(1)dy/dx]/(2-xy)^2

d/dx (3y^2)=(3x^2)(2-xy)+x^3y+x(dy/dx)]/(2-xy)^2

(2-xy)^2(3y^2)(dy/dx)-x (dy/dx)=3x^2(2-xy)+x^3y
dy/dx[(2-xy)^2(3y^2)-x]=3x^2(2-xy)+x^3y
dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)-x]

is this correct now?

4. Look at this part right here for now:
$\displaystyle -x^3(-y-xdy/dx)$

$\displaystyle x^3y+xdy/dx$

Are you sure?

5. Originally Posted by eXist
Look at this part right here for now:
$\displaystyle -x^3(-y-xdy/dx)$

$\displaystyle x^3y+xdy/dx$

Are you sure?

so should it be this..
final result:

dy/dx=[3x^2(2-xy)+x^3y]/(2-xy)^2(3y^2)-x^4

?correct now?

6. Looks good now ! Just to make sure others understand, make sure to use parenthesis on the bottom:

Yours:
dy/dx=[3x^2(2-xy)+x^3y]/(2-xy)^2(3y^2)-x^4

Yours with parenthesis/brackets:
dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)-x^4]

I'm sure that's what you meant, seeing how your work is correct now.